Let E_n= $\frac{-1\mathit{m}{\mathit{e}}^4}{8{\mathit{\epsilon }}_0^2{\mathit{n}}^2{\mathit{h}}^2}$ be the energy of the nth level of H atom. If all the H-atoms are in the ground state and radiation of frequency $\frac{{\mathit{E}}_2-{\mathit{E}}_1}{\mathit{h}}$  falls on it

(a) It will not be absorbed at all

(b) Some of atoms will move to first excited state

(c) All atoms will be excited to the n= 2 state

(d) No atoms will make a transition to then n=3 state

Kindly go through the solution 

Change in surface energy = work done

$\left|\Delta E_0\right|=\left(-136\left\{1-\frac{1}{4}\right\}\right)eV$

|DE₀| = –10.2

$\lambda =\frac{12400}{10.2}\times 10^{-10}m$

$\rho =\frac{h}{\lambda }=\frac{6.63\times 10^{-34}\times 10.2}{12400\times 10^{-10}}$                  

$?$ $mv=\frac{h}{\lambda }$            

  $1.8\times 10^{-27}$           

$v=\frac{6.63\times 10.2\times 10^{-34}}{12400\times 10^{-10}}$

$v=\frac{6.63\times 10.2}{12400\times 1.8}\times 10^3$            

$=\frac{6.63\times 102}{124\times 1.8}=3.02$ ]

= 3 m/s

$-0.85=\frac{-13.6}{n^2}$  

n = 4

Number of transitions = $\frac{4\times 3}{2}=6$  

Kinetic energy: Potential energy = 1 : –2